In this post of mine on the
I found a general formula for the arc length of the link name above. This formula cannot cancel the radicals and radicand as this may get rid of its power to find results for $\mathrm {x,n}\not>0$. Incomplete Beta function and falling factorial $\ (x)_y$:
$$\mathrm{Arclength(x^n) \text{ with respect to x=}\frac{xB_{-n^2x^{2n-2}}\left(\frac1{2n-2},\frac32\right)}{2\sqrt[2n-2]{-n^2x^{2n-2}}(n-1)}}$$
$$\mathrm{\text{ arc length of } x^i \text{ wrt x from 1 to 2}= -\frac{1+i}4 B_{2^{2i-2}}\left(-\frac{1+i}4,\frac32\right)-\frac{\sqrt\pi}2\left(-\frac{1+i}{4}\right)_{-\frac12}= -\frac{1+i}4 B_{\frac14\left(cos(2ln(2))+isin(2ln(2))\right)}\left(-\frac{1+i}4,\frac32\right)-\frac{\sqrt\pi}2\frac{Γ\left(\frac{3-i}4\right)}{Γ\left(\frac{5-i}4\right)}=.780647…+.168809…i}$$
Taking the real and imaginary parts gives graph:
$$\mathrm{x^i=e^{iln(x)}=cos(ln(x))+i\ sin(ln(x))\mathop \implies Arclength(cos(ln(x)))=Re\left(-\left(x^{2i-2} \right) _{\frac{1+i}{4}} \frac{1+i}{4}B_{x^{2i-2}}\left(-\frac{1+i}4,\frac32\right)\right), Arclength(sin(ln(x)))=Im\left(-\left(x^{2i-2} \right) _{\frac{1+i}{4}} \frac{1+i}{4}B_{x^{2i-2}}\left(-\frac{1+i}4,\frac32\right)\right)}$$
I want to simplify the arc length from 1 to 2 as I do not think the arc length from 0 to 1 converges:
$$\mathrm{Arclength(cos(ln(x)))\text{ from 1 to 2 }=Arclength(sin(ln(x)))\text{ from 1 to 2} = \int_1^2 \sqrt{ \frac{sin^2(ln(x))}{x^2}+1}dx= Re\left(-\frac{1+i}4 B_{2^{2i-2}}\left(-\frac{1+i}4,\frac32\right)\right)-\frac{\sqrt\pi}2 Re\left(\frac{Γ\left(\frac{3-i}4\right)}{Γ\left(\frac{5-i}4\right)}\right)}$$
$$\mathrm{ Arclength(sin(ln(x)))\text{ from 1 to 2} =\int_1^2\sqrt{\frac{cos^2(ln(x))}{x^2}+1}dx=Im\left(-\frac{1+i}4 B_{2^{2i-2}}\left(-\frac{1+i}4,\frac32\right)\right)-\frac{\sqrt\pi}2 Im\left({\frac{Γ\left(\frac{3-i}4\right)}{Γ\left(\frac{5-i}4\right)}}\right)}$$
Here is how to find the
Based on the logic of finding the real and imaginary parts of an arclength, one can get 2 new arc length results and area under arc length curves. The problem boils down to finding real and imaginary parts of a falling factorial and incomplete beta function. Note that there is no known closed form for the real and imaginary parts of Γ(i), but if you can find a closed form or series solution to the real and imaginary parts, this will also work. Please correct me and give me feedback!
